Question

If $C_0,C_1, C_2, ....,C_n$ are binomial coefficients of order n, then the value of $\frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+...=$

• A. $\frac{2^n +1}{n+1}$
• B. $\frac{2^n -1}{n+1}$
• C. $\frac{2^n +1}{n -1}$
• D. $\frac{2^n }{n+1}$

Answer 1

Answer: (B)

$\frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+...=\frac{n}{1\cdot 2}$
$+\frac{ n\left(n-1\right)\left(n-2\right)}{1\cdot 2\cdot 3\cdot 4} \frac{+n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}+.....$
$ = \frac{1}{\left(n+1\right)}\left[\frac{n\left(n+1\right)}{2!}+\frac{n\left(n-1\right)\left(n-2\right)\left(n+1\right)}{4!}+...\right]$
$=\frac{1}{n+1}\left[^{n+1}C_{2}+ \,{}^{n+1}C_{4} +......\right]$
$ =\frac{1}{n+1} \left[2^{n+1-1}-1\right] =\frac{2^{n }-1}{n+1}$